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Step-by-Step Solution
Step 1: Write down the photoelectric effect equation
According to Einstein’s photoelectric equation, the energy of the incident photon is used partly to overcome the work function (threshold energy) of the metal and the rest appears as the maximum kinetic energy (related to the stopping potential). Mathematically:
$ \frac{hc}{\lambda} = \phi + e V_s $
where:
$ h $ is Planck’s constant.
$ c $ is the speed of light.
$ \lambda $ is the wavelength of the incident radiation.
$ \phi $ is the work function of the material.
$ e $ is the charge of an electron.
$ V_s $ is the stopping potential.
Step 2: Express the equations for two different wavelengths
Let $ \lambda_1 = 491 \text{ nm} $ and $ \lambda_2 $ be the new wavelength to be found. The corresponding stopping potentials are $ V_{s1} = 0.710 \text{ V} $ and $ V_{s2} = 1.43 \text{ V} $, respectively. Then we have two equations:
$ e V_{s1} = \frac{hc}{\lambda_1} - \phi \quad (1) $
$ e V_{s2} = \frac{hc}{\lambda_2} - \phi \quad (2) $
Step 3: Subtract the two equations
Subtract equation (1) from equation (2) to eliminate the work function $ \phi $:
$ e V_{s2} - e V_{s1} = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} $
or
$ e \left( V_{s2} - V_{s1} \right) = hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) $
Substitute the given values: $ V_{s1} = 0.710 \text{ V}, V_{s2} = 1.43 \text{ V}, \lambda_1 = 491 \text{ nm}, \lambda_2 = ? $.
Step 4: Substitute numerical values and simplify
Commonly, for calculations involving eV and nm, a useful approximation is $ \frac{hc}{e} \approx 1240 \text{ eV·nm} $.
So, from
$ e (1.43 - 0.710) = 1240 \left( \frac{1}{\lambda_2} - \frac{1}{491} \right), $
we get
$ (1.43 - 0.710) = 1240 \left( \frac{1}{\lambda_2} - \frac{1}{491} \right). $
Thus,
$ 0.72 = 1240 \left( \frac{1}{\lambda_2} - \frac{1}{491} \right). $
Rearrange and solve for $ \frac{1}{\lambda_2} $:
$ \frac{0.72}{1240} = \frac{1}{\lambda_2} - \frac{1}{491}. $
$ \frac{1}{\lambda_2} = \frac{1}{491} + \frac{0.72}{1240}. $
Step 5: Calculate the new wavelength $ \lambda_2 $
First, compute each term:
$ \frac{1}{491} \approx 0.00203 \text{ nm}^{-1} $
$ \frac{0.72}{1240} \approx 0.00058 \text{ nm}^{-1} $
So,
$ \frac{1}{\lambda_2} = 0.00203 + 0.00058 = 0.00261 \text{ nm}^{-1}. $
Therefore,
$ \lambda_2 = \frac{1}{0.00261} \approx 383.14 \text{ nm}. $
Rounding suitably gives $ \lambda_2 \approx 382 \text{ nm}. $
Step 6: State the final answer
Hence, the new wavelength of the incident light is approximately 382 nm.
